Question

Write the expression as the logarithm of a single quantity. $$ \frac{1}{3}\left[2 \ln (x+3)+\ln x-\ln \left(x^{2}-1\right)\right] $$

Short answer

\[\ln \left(\frac{(x+3)^2x}{(x^{2}-1)}\right)^{\frac{1}{3}}\]

Step-by-step solution

Apply Power Rule of Logarithms

Start by applying the first property of logarithms mentioned above. This is only applicable to the first term in the given expression, because it's the only term where the coefficient of the logarithm isn't 1. This results in: \[\frac{1}{3}\left[\ln ((x+3)^2)+\ln x-\ln \left(x^{2}-1\right)\right]\]

Combine Logarithms Into One

The next step is to use the rule of adding and subtracting logarithms which states that the addition of two logarithms is the logarithm of their multiplication and the subtraction of two logarithms is the logarithm of their division. Apply these rules to the expression we now have: \[\frac{1}{3}\ln \left(\frac{(x+3)^2x}{(x^{2}-1)}\right)\]

Eliminate Fraction From Expression

Lastly, use the first property again, this time in reverse. This will eliminate the fraction in front of the logarithm, leaving us with a single logarithm. This results in: \[\ln \left(\frac{(x+3)^2x}{(x^{2}-1)}\right)^{\frac{1}{3}}\]